91Ó°ÊÓ

Use torque equations to find the applied force and the tangential force applied to grindstone.

EXECUTIVE:

Given that the mass of grid stone,

$\mathrm{m}=50.0\mathrm{kg}$

Diameter of the grid stone, $d=0.520\mathrm{m}$

Radius of the grid stone,localid="1667793725952" $r=\frac{d}{2}$

$=\frac{d}{2}$

$$\begin{gathered} =\frac{0.520m}{2} \\ =0.26m \end{gathered}$$

Moment of inertia of the solid disk,

$$\begin{gathered} I=\frac{1}{2m{r}^2} \\ =\frac{1}{2}(50.0kg)(0.26m{)}^2 \\ =1.69kg.m^2 \end{gathered}$$

Normal force acting between grid stone and the ax,

$F_N=160N$

Coefficient of kinetic friction between the blade and the grid stone,

$\mathrm{μ}=0.60$

Frictional force acting between the grid stone and the ax is

$$\begin{gathered} f=\mathrm{μ}{F}_N \\ =(0.60)\left(160N\right) \\ =96N \end{gathered}$$

Frictional torque acting between the axle of the stone and the bearing,

${\mathrm{Ï„}}_f=6.50N.m$

Length of the crank handle, $L=0.500m$

Initial angular velocity of the grid stone,

${\mathrm{Ó\neg }}_1=0 \mathrm{rad}/\mathrm{s}$

Final angular velocity of the grid stone,

${\mathrm{Ó\neg }}_2=120rev/min$

$$\begin{gathered} =(120rev/min)\frac{2\mathrm{π}rad}{1rev}\left(\frac{1 min}{60s}0\right) \\ =12.57 rad/s \end{gathered}$$

Time taken $t=9.00s$

From rotational kinematic equations angular acceleration of the grid stone

localid="1667794528792" $$\begin{gathered} \mathrm{Î\pm }=\frac{{\mathrm{Ó\neg }}_2-{\mathrm{Ó\neg }}_1}{t} \\ =\frac{12.566"rad/s-0}{9.00 s} \\ =1.396 rad/s^2 \end{gathered}$$

Net torque acting on the grid stone is

$$\begin{gathered} \mathrm{Ï„}=I\mathrm{Î\pm } \\ LF-{\mathrm{Ï„}}_F-rf=I\mathrm{Î\pm } \\ LF=I\mathrm{Î\pm }+{\mathrm{Ï„}}_f+rf \\ F=\frac{I\mathrm{Î\pm }+{\mathrm{Ï„}}_1+rf}{L} \\ =\frac{(1.69kg.m^2)(1.36rad/s^2)+6.50"N.m+"(0.26m)\left(96N\right)}{0.500 m} \\ =\boxed{67.6N } \end{gathered}$$

Hence, the force applied is $67.6 N$.

As the grid stone moving with constant angular speed, angular acceleration of the wheel is zero. Hence the net torque acting on the grid stone is zero.

$$\begin{gathered} \mathrm{τ}=0 \\ LF-{\mathrm{τ}}_f-rf=0 \\ LF={\mathrm{τ}}_f+rf \\ F=\frac{{\mathrm{τ}}_f+rf}{L} \\ =\frac{6.50N.m+(0.26m)\left(96N\right)}{0.500m} \\ =\boxed{62.9 N} \end{gathered}$$

Hence, the net torque is $=62.9 N$.

Angular velocity of the grid stone,

$$\begin{gathered} {\mathrm{Ó\neg }}_3=120rev/min \\ =(120 rev/min)\left(\frac{2\mathrm{Ï€} rad}{1rev}\right)\left(\frac{1min}{60 s}\right) \\ =12.57 rad/s \end{gathered}$$

Final angular velocity of the grid stone,

${\mathrm{Ó\neg }}_4=0 rad/s$

Angular acceleration of the grid stone from rotational kinematics

$\mathrm{Î\pm }=\frac{{\mathrm{Ó\neg }}_2-{\mathrm{Ó\neg }}_1}{t}$

Torque acting on the grid stone due to axle is

localid="1667795554159" $$\begin{gathered} {\mathrm{Ï„}}_f=I\mathrm{Î\pm } \\ {\mathrm{Ï„}}_f=I\left(\frac{{\mathrm{Ó\neg }}_4-{\mathrm{Ó\neg }}_3}{t}\right) \\ t=I\left(\frac{{\mathrm{Ó\neg }}_4-{\mathrm{Ó\neg }}_3}{{\mathrm{Ï„}}_f}\right) \\ =(1.69kg.m^2)\left(\frac{0-12.57rad/s}{-6.5N.m}\right) \\ =\boxed{3.27s} \end{gathered}$$

Hence, torque acting on the grid stone due to axle is .$3.27s$